2017 AIME II Problems/Problem 13
Contents
Problem
For each integer , let be the number of -element subsets of the vertices of the regular -gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of such that .
Solution 1
Considering , we have the following formulas:
:
:
:
:
To derive these formulas, we note the following: Any isosceles triangle formed by the vertices of our regular -sided polygon has its sides from the set of edges and diagonals of . Notably, as two sides of an isosceles triangle must be equal, it is important to use the property that same-lengthed edges and diagonals come in groups of , unless is even when one set of diagonals (those which bisect the polygon) comes in a group of . Three properties hold true of :
When is odd there are satisfactory subsets (This can be chosen with choices for the not-base vertex, and for the pair of equal sides as we have edges to choose from, and we must divide by 2 for over-count).*
- Another explanation: For any diagonal or side of the polygon chosen as the base of the isosceles triangle, there is exactly 1 isosceles triangle that can be formed. So, the total number of satisfactory subsets is
When is even there are satisfactory subsets (This can be chosen with choices for the not-base vertex, and for the pair of equal sides as we have edges to choose from, one of them which is not satisfactory (the bisecting edge), and we must divide by 2 for over-count).
When is a multiple of three we additionally over-count equilateral triangles, of which there are . As we count them three times, we are two times over, so we subtract .
Considering the six possibilities and solving, we find that the only valid solutions are , from which the answer is .
Solution 2 (elaborates on the possible cases)
In the case that , there are equilateral triangles. We will now count the number of non-equilateral isosceles triangles in this case.
Select a vertex of a regular -gon. We will count the number of isosceles triangles with their vertex at . (In other words, we are counting the number of isosceles triangles with among the vertices of the -gon, and .)
If the side spans sides of the -gon (where ), the side must span sides of the -gon, and, thus, the side must span sides of the -gon. As has three distinct vertices, the side must span at least one side, so . Combining this inequality with the fact that and (as cannot be equilateral), we find that there are possible .
As each of the vertices can be the vertex of a given triangle , there are non-equilateral isosceles triangles.
Adding in the equilateral triangles, we find that for : .
On the other hand, if , there are no equilateral triangles, and we may follow the logic of the paragraph above to find that .
We may now rewrite the given equation, based on the remainder leaves when divided by 3.
Case 1:
The equation for this case is .
In this case, is of the form or , for some integer .
Subcase 1: Plugging into the equation above yields .
Subcase 2: Plugging into the equation above yields , which has no integer solutions.
Case 2: The equation for this case is .
In this case, is of the form or , for some integer .
Subcase 1: In this case, the equation above yields .
Subcase 2: In this case, the equation above yields .
Case 3: The equation for this case is .
In this case, is of the form or , for some integer .
Subcase 1: The equation above reduces to , which has no integer solutions.
Subcase 2: The equation above reduces to , which does not yield a positive integer solution for .
In summary, the possible are , which add to .
Solution 3
We first notice that when a polygon has sides where , there cannot exist any three vertices that form an equilateral triangle. Also, the parity of and also matters, since they influence how many isosceles triangles including equilateral triangles exist in the polygon. We can model an equation , where the lines that are congruent connect the vertices that are vertices apart and the other line has vertices that are vertices apart. If is even, there are solutions for which would create a unique type of isosceles triangle. We subtract two since cannot be zero. If is odd, there are solutions for . Next, we do casework on the congruence of and the parody of using the information above:
Case 1:
,
is even, is odd
There are unique types of isosceles triangles in the polygon with sides. Each isosceles triangle has a unique point which connects the two congruent sides. Therefore, for each type of isosceles triangle, there exists of those triangles since the unique point can be any of the vertices. There are types of isosceles triangles in the polygon with sides and unique points for each type of triangle. Therefore, . Solving, we get .
Case 2:
,
is odd, is even
Using similar logic, there are possible isosceles triangles in the sided polygon. There are possible isosceles triangles in the sided polygon. The difference should be , so . Solving gives .
For both of these cases, we don't have to worry about equilateral triangles since in both cases .
Case 3:
,
is even, is odd
There are possible isosceles triangles in the sided polygon. The case is a bit more complicated, as we have to consider equilateral triangles as well. In this case, there is one solution where , which would produce an equilateral triangle. Therefore, we subtract that case to calculate only isosceles triangles with 2 congruent sides. Only isosceles triangles with exactly 2 congruent sides have a unique point, while there exist only distinct equilateral triangles in a polygon with sides, the rest are equivalent and symmetrical. Therefore, there are isosceles triangles with at least 2 congruent sides in a polygon with sides. Putting it together, . Solving yields which is impossible since has to be an integer. Therefore, this case is not valid.
Case 4:
,
is odd, is even
There are isosceles triangles in the sided polygon. Using the idea above, there are isosceles triangles with 2 congruent sides in an sided polygon. There are equilateral triangles. Therefore, . Looking at the left hand side, it is clear that has to be negative, which is not valid. Therefore, this case is not valid.
Case 5:
,
is even, is odd
There are a total of isosceles triangles in a polygon with sides. There are a total of isosceles triangles in a polygon with sides. Therefore, . Simplifying, we get .
Case 6:
,
is odd, is even
There are a total of isosceles triangles in the polygon with sides. There are isosceles triangles in the polygon with sides. Therefore,. Simplifying, we get , which means is not an integer. Thus, this case is invalid.
Adding all the valid cases, we obtain
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See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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